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# lecture_11 - Lecture 11 0.1 Forced response Very often the...

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Unformatted text preview: Lecture 11 0.1 Forced response Very often, the force input into a spring-damper-mass system is not periodic. The earthquake vibration of the ground is not, and neither is the force that stresses your cell phone when it hits the ground. We do not like our gadgets to break when something like that happens and so engineers that are involved in design of electronic devices spend a lot of time making sure that repeated impacts do not harm the basic func- tionality of the device. This is why the Wii controller that I discussed in the last lecture has a silicon sleeve jacket. For arbitrary inputs, we use an approach that superposes many impulsive forces over the period of time that the force is acting. We will use the so-called Dirac δ function, which is the limit of functions (see figure 1) f = 1 < t ≤ (1) f = 0 t > (2) as goes to zero. Note that the integral of f is always 1, no matter what is. Let’s consider a mass that is acted upon by such a force. By Figure 1: Function f . 1 Newton’s second law m ¨ x = f ( t ) and thus, by integrating over time interval [0 , ] we get Z m ¨ x ( ) dt = Z f ( t ) dt = 1 and thus m ˙ x ( )- m ˙ x (0) = 1 . Observing the zero initial velocity condition ˙ x (0) = 0, we get ˙ x ( ) = 1 m . Also, lim → x ( ) = x (0) , so it will be ∞ except if x = 0 . Physically, this means that under an impulsive force of unit magnitude (integral) the linear momentum of a free mass m changes to m ˙ x (0 + ) = 1 where 0 + denotes the moment of time right after the impulsive force action, while its position does not change, i.e. x (0 + ) = 0. Let’s now consider the spring-mass system impacted by the force f : m ¨ x + c ˙ x + kx = δ ( t ) . If we integrate the above equation from t = 0 to t = , we get: m (˙ x ( )- x (0)) + c ( x ( )- x (0)) + k Z xdt = 1 Since c ( x ( )- x (0)) = 0 and k Z xdt = 0 we have m (˙ x ( )- x (0)) = 1 ....
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lecture_11 - Lecture 11 0.1 Forced response Very often the...

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